Midyear Review Solutions

The solutions are posted below - I have updated some of the previous answers with a bit more details, so be sure to check here for the latest answers.

Please also post a comment to this blog post on what topics/problem types you personally want the most help with during class tomorrow.

These are the problems I have created for tomorrow - I suggest you try them and come in with questions: Review problems for Monday

Midyear Review #1 - 1D kinematics
1.
(b) slope is zero, zero acceleration
(c) zero
(d)&(e) 200 meters
2.
(b) 8 m/s^2
(c) area under v vs. t graph, distance = 100 m.
(d) Y = vo*t + .5at^2 = 100 m.
(e) 16 m/s
(f) graph should be a parabola with its vertex at t = 0, y = 0.

3.
(a) 4 seconds
(b) horizontal line from t = 0 to t = 3, then a diagonal line going down to zero.
(c) 60 meters
(d) 100 meters

4. 24 m/s

5.
(a) +15 m/s, -10 m/s
(b) change in velocity is -25 m/s (or 25 m/s directed upwards)
(c) 50 m/s^2 directed upwards

6.
(a) 4 m/s^2
(b)East
(c) velocity directed south, acceleration directed East

Midyear Review #2 - 2D kinematics and Projectile Motion:
1.
(a) 1.563 seconds
(b) 7.813 meters
(c) 13.53 meters

2.
(a) 4.0 seconds
(b) 5 m/s
(c) (We can assume that the net force is mostly vertical since the horizontal velocity (5 m/s) is so much less than the vertical velocity (40 m/s). acceleration = .769 m/s^2, displacement = 1040 meters (that's a deep target!)

3.
(a) 412.3 km at 14 degrees North of East.
(b) 42.7 km/h
(c) 51.7 km/h

Midyear Review #3 - Newton's Laws


1. 4 N directed upwards

2. T1 (diagonal cord) = 200 N, T2 (horizontal cord) = 173 N

3.
(a) a = F/6m
(b) Block 1 (3m), Block 2 (m), Block 3 (2m)
Fnet1 = F/2
Fnet2 = F/6
Fnet3 = F/3
(c) T1 (between 3m and m) = F/2, T2 (between m and 2m) = 2/3F


4. The key thing to think about is that WALL-E always hangs from EVE's arms. The FBD of WALL-E always has EVE's force directed upwards, and his weight directed downwards. The difference is the direction of acceleration, which must always be directed towards the center of the circle.

For (a), the acceleration is upwards, so F – mg = mv^2/R. F = 669 N.

For (b), the acceleration is downwards, so mg – F = mv^2/R. F = 81 N.

5.
(a) T1 = 60 N, T = 100 N, minimum value of static friction coefficient = .27. Notice that this is the solution if you say that static friction is directed DOWN the inline.

If you instead make the static friction force go UP the incline in your FBD, you get a negative value for mu, which means that this is NOT possible in this situation.

(b) a = 1.31 0.670 m/s^2, T = 84.9 91.3 N. Notice that it says to assume the incline is smooth, which means no kinetic friction.

c) If the cord is cut, the right block will fall at g, or 9.8 m/s^2. The left block will slide at g sin(60) = 8.49 m/s^2. The answers must be different because the two blocks are no longer linked by the cord.

Midyear Review #4 – Circular Motion, Gravity, SHM

1. Be sure to use the static friction coefficient, as static friction is what keeps the car from slipping off the road. Kinetic friction occurs when two surfaces slide past each other.

Max speed = (mu*g*r)^(1/2) = 23.7 m/s

2.
(a) Mass is the same, 70 kg
(b) 700 N (16) = 11,200 N
(c)160 m/s^2


3. Remember to draw a FBD of the satellite, with the gravity force directed towards the Earth's center, and positive also directed towards the center. You should get that v = (GM/r)^(1/2).

4.use your answer to #3 to justify your answer here. Also, notice that the mass does NOT change your answer. The ratio is equal to (1/5)^(1/2)

5.
a) Since the mass is not moving when the spring is at maximum compression, the KE = 0.

b) U = 1/2*k*delta_x^2, so the potential energy = 1/9*maximum U = 8 J.

c) Since the total energy is constant (energy is conserved!), the KE = 64 J.

d) T = 2*pi(m/k)^(1/2), but since you aren't given the mass in this problem, you can't solve it.

MARLON: You're slacking!

I know, I know. So let's suppose the mass was 1 kg. This would mean the period would be 0.363 seconds.

e) If the pendulum has the same frequency as the mass on the spring, it must also have the same period. The period of oscillation for a pendulum is given by T = 2*pi(l/g)^(1/2) so for a period of 0.363 seconds, this means the length would have to be 3.27 cm. That's a small pendulum, so it will swing back and forth fairly quickly.

MARLON: That pendulum is NOT slacking!

I couldn't agree with you more.

6.
speed, max D, min A
kinetic energy max D, min A
potential energy, max A, min D
total energy, constant
angular momentum, constant
net force on the satellite, max D, min A
acceleration, max D, min A

Capacitors? Yes you CAN!

Reminder: The quizam will have a mixed bag of electrostatics AP problems, and might have another easy-to-predict surprise....

here's a hint:

12:41 PM, January 20, 2009

Solutions: Chapter 17, Q13, P32, 34, 37, 39, 47

Questions:

Problems:
32. Q = CV, so V = 22.0 volts.

34. using the equation for a parallel plate capacitor: C = e₀*A/d so the area = 5.0 x 10⁷ m^2. That's a lot of area!

37. We can get the electric field using V = Ed. Also, Q = CV.

We also need the fact that for a parallel plate capacitor, C = e₀*A/d.

We can put ALL of this together to get that Q = e₀*A*E*d/d = e₀*A*E , all of which we know! Substituting, Q = 2.63 x 10^-8 C.

39. Q = CV, so V = 90 volts. The potential difference V = Ed, so E = V/d = 4.5 x 10⁴ V/m or N/C.

47. The key to these problems is to use the equation in the form that does NOT change. For part (a), for example, it does NOT make sense to use U = 1/2*QV because the charge Q depends on the voltage through Q = CV. If you instead use U = 1/2CV², C is a constant, and V is the only thing that changes.

a) The energy must double, since U = 1/2CV² and C is a constant.

b) In this case, the Q is changed, and C must stay constant. We can then see by U = Q²/2C that the energy must be multiplied by a factor of 4.

c) If the battery stays connected, it is the VOLTAGE that stays the same, but the capacitance is divided by 2. Since U = 1/2CV^2, the energy will be divided by 2.

How much for a charged particle in mint condition?

Definition quiz Friday. Word.

Click here for the solutions to the problems from class.

From Infinity...to a distance R!

For those of you reading the blog, just a heads up that tomorrow IS Wednesday, our day for an AP problem quiz. Another bit of useful info...it will HEAVILY be based on kinematics. Might be a good idea to check out those review problems on kinematics handed out in class - the solutions are posted on the blog in January 2008. Use the menu on the right hand side to find those solutions....

Now for the solutions:

p. 522, 11, 12, 15, 16, 18, 20, 23

11. KE = 1/2*mv^2 with KE = 28 x 10⁶ eV * (1.6 x 10^-19 J/eV) = 4.5 x 10^-12 J. v = (2*KE/m)^(1/2) = 7.3 x 10⁷ m/s.

12. The same procedure, except we use the mass of the alpha particle. v = 1.6 x 10⁷ m/s.

15. We must first find the total potential at each location. The first location has a distance of 16 cm from both charges. Using V = kQ/R and adding the potential for each charge, the total V in the middle is 3.38 x 10⁶ V.

In the second position, the distance from one charge is 6 cm, and the other is 26 cm. We again use V = kQ/R and add the two results together to get 5.54 x 10⁶ V.

We can neglect the change of kinetic energy during this process because the charge is being 'placed' at each location.

Wncf = delta_K + delta_U = q*delta_V = 1.08 J

16.
a) The electric potential of the proton is V = ke/R = 5.8 x 10⁵ V

b) The potential energy of the system is found by multiplying the potential created by the proton (the answer from a) by the proton located at the given distance. Since V = U/q, U = q*V = 9.2 x 10^-12 J.

18. This follows the same pattern as what we did at the end of today.

The first electron takes zero work to bring together.

The second electron takes e*delta_V = -e*(k(-e)/R - 0) = ke^2/R.

The third electron must overcome the repulsive forces of BOTH electrons that are already there: e*delta_V = -e*(k(-e)/R - 0) - e*(k(-e)/R - 0)= 2ke^2/R.

The total must then be the sum of these quantities, 3*ke^2/R = 6.9 x 10^-18 J.

20. Energy is conserved, so:
initial PE + initial KE = final PE + final KE
ke*Q/R + 0 = 0 (since it is at infinity) + 1/2*mv^2
v = 2.33 x 10⁷ m/s.

23. We first need to find the total potential at each location.

At A:
V_A = kq/b + k*(-q)/(d - b) which can be simplified to kq(d - 2b)/b(d-b).

At B:
V_B = kq/(d - b) + k*(-q)/b = kq(2b - d)/b(d - b).

The difference in potential Vb - Va can be simplified to 2kq(2b - d)/b(d-b).

You have a lot of POTENTIAL...but you need to WORK harder.

p. 522, Q7, P 4,5,6,8,9

Question 7. Electric potential is the amount of energy per unit charge, while electric field is the amount of electric FORCE per unit charge. Electric potential is just electric potential energy divided by charge.

Problems.
4. If the electron gains 3.6 x 10^-16 J of KE, it means it loses a potential energy of 3.6 x 10^-16 J. This happens as it crosses a potential difference delta_V = delta_U/q.

Thus delta_V = (3.6 x 10^-16 J)/(-1.6 x 10^-19 C) = 2,156 V.

Since it is an electron with a negative charge, the final plate (plate B) is at the higher potential.

5. delta_V = Ed, E = delta_V/d = 220 V/(.0052 m) = 4.23 x 10⁴ N/C or V/m.

6. Following the same procedure as with #5, the answer is 7.04 V.

8. d = delta_V / E = 3.3 x 10^-5 m

9. Since we have a non-conservative force doing work, a change in kinetic energy, and potential difference, it sounds like we might be able to use the generalized work energy theorem.

W_ncf = delta_K + delta_U
25 x 10^-4 J = 1.82 x 10^-4 J + delta_U
delta_U = 2.02 x 10^-3 J
delta_V = delta_U/q = 2.02 x 10^-3 J/(-7.5 x 10^-6 C = -269 V. This is the potential difference from a to b for the negative charge. The potential difference Va - Vb (the reverse of Vb - Va) = +269 V.

Charging by Induction, Conduction, Frustration

HW: Q7,8, P19, 27, 45, 56
Questions:

7. The attraction of the negative charges to the positive charges is strong, but remember that the charges will also feel a repulsion force from being close to the other negative charges.

8. The leaves of the electroscope act like springs as they are forced apart. You can feel this force in a simple experiment - take a ruler and hold it against the edge of a table so that most of its length is hanging off the edge. As you push on the free end of the ruler, you will feel a force that is proportional to how far you push on the end (displacement).

Problems:
19. To figure out where the third charge must go, you have to observe that it can't be located between the charges, because the forces from each charge would be pointed in the SAME direction. You can then say that the third charge is located a distance x to the right of the negative charge.

F1 = kQ1Q/r1^2
F2 = kQ2Q/r2^2

substituting for r1, r2, and dividing by k:
Q1/(L + x)^2 = Q2/x^2

You can solve numerically, getting that x = .91 m or -.11 m. Since the -.11 m location puts the charge between the first two (which we previously said could not happen), only the x = .91 m solution works.

Important Information Alert!
Notice that in 27, 45, and 56, it is stated in the problem that electric field is constant. If you skim through the problem too quickly, you may miss this fact.

27. F = qE = ma so E = 7.10 x 10^-10 N/C. Since it is an electron, the electric field will be directed in the opposite direction of the acceleration, which would be to the south.

45. The weight of the water drop is balanced by the electric force.

mg - qE = 0.

To find m, you must use the volume of a sphere, 4/3*pi*r^3, times the density of water, 1000 kg/m^3.

You can then solve for Q, and then divide this value by the charge on an electron (1.6 x 10^-19 C) to find the number of electrons on the drop.

56. Draw a FBD: T + qE - mg = 0. Thus E = (mg - T)/Q = 1.06 x 10⁷ N/C. Since we assumed in our FBD that E was directed upwards, the negative answer means the field is actually directed downwards.

There Is No Electric Field inside Tom - He's the Conductor!

Electric Field solutions:

p 496 - 498, Q 18, 19, P 33, 34, 35, 38, 43

1981B3 Solutions:
b) T cos(th) - mg = 0. T = mg/cos(th) = 5.65 x 10^-2 N

T sin(th) - qE = 0. E = T sin(th)/Q = 5650 N/C

c) Since the electric field will accelerate the sphere in the horizontal direction, and gravity will accelerate the ball in the vertical direction, the sphere will travel in a straight line directed downwards.

Questions:

18. The field decreases in strength A, B, and then C.

19. If field lines crossed, it would mean that a positive charge could move in two possible directions at the intersection of the lines. This contradicts the definition of the direction of electric field that we came up with, which was THE direction that a positive charge would move at a location.

33.

34.

35. Draw a FBD: Fnet = qE = ma. You know the mass of a proton from the back of the book, as well as the charge, so you can solve for E = 0.10 N/C.


38. The acceleration from the field comes from qE = ma, such that a = 3.24 x 10¹⁵ m/s^2.

We can use our constant acceleration equations because E, and therefore the electric force are constant. v^2 = vo^2 + 2a*delta_x.

v = 8.83 x 10⁶ m/s

b) The ratio of the forces qE/mg = 3.0 x 10^-15.

43. Again, from equilibrium, qE - mg = 0. E = (1.67 x 10^-27 kg)(9.8 m/s^2)/(1.6 x 10^-19C = 1.02 x 10^-7 N/C directed upwards.

Charging Into The Field - Electric field and 2D Electrostatics

Problems 12, 13, 18, 22, 26, 29

12. Remember the trick using your thumbs to find the direction of the force - same distance for all three, and a 60 degree angle for each force. Don't forget to find the direction - 83.8 N directed AWAY from the center of the triangle.

13. Again, this is not actually as tedious as it could be since it is a square and the charges are all the same, so your answer will end up being the same for each charge. You must use the Pythagorean theorem to find the distance of each charge from the charge on the opposite corner of the square (1.41 m).

For each charge, you will end up with something like Fnetx = F1 + F2 sin 45, and Fnety = F3 + F2 cos 45. (The subscripts will be different for each charge.) The final answer should be 6.2 x 10⁵ N directed away from the center of the square.

18. A little mix of Newton's 2nd and electrostatics here - the FBD of the electron should have kQ1Q2/r^2 directed towards the center of the circular orbit. The acceleration is v^2/R.

Q1 = Q2 = e (the fundamental charge), so ke^2/R^2 = mv^2/R.

Thus R = ke^2/mv^2 = 2.2 x 10^-10 m. It is interesting that this is not actually so far off from the actual size of an atom - you have now used basic physics to derive this fact! Not bad, eh?

22. This is a basic application of the definition of electric field: E = F/q. The magnitude comes from rearranging this definition: F = qE = 5.6 x 10^-16 N. Since it is an electron, which has a negative charge, the force will be directed towards the West, opposite the field.

26. This problem is very much like the one we did in class during 7th period - pretend that there is a small test charge at the midpoint of the two charges. Both E1 and E2 are directed to the left (or away from the positive charge) so their magnitudes should be added together. The net field (using the electric field for a point charge E = kQ/R^2) is 3.2 x 10⁸ N/C.

29. We need to use the SAME concept from the other problems in 2D to solve this problem. Unfortunately it's all algebraic, but fear not. The key is to start with a good diagram. I'm going to take mine from the nice folks at the textbook company:


In the diagram, E_- represents the electric field direction from the negative charge, and the other represents the direction for the positive field.

The distance from the origin is x, so the distance from x to each charge is (x^2 + a^2)^(1/2).

The other trick is to find the angle, and THEN find the sine of the angle, but you don't actually have to break that up into two steps. We know from trigonometric ratios that Sin (theta) = opposite side/hypotenuse. The hypotenuse is (x^2 + a^2)^(1/2) and the opposite side of the angle in the right triangle drawn by E_- is a. this means that sin(theta) = a/(x^2 + a^2)^(1/2).

The net electric field in x will be zero, since the charges are the same. In the y-direction, you should get that E = 2*E_- sin(theta).

Substituting our expression for the sine and simplifying, E = 2kQa/(a^2 + x^2)^3/2

But Officer, You Haven't Charged me with Anything! - Electrostatics

P. 496 - 497: Q1, P1, 2, 8, 11, 20

Q1.

Problems:
1. 1.88 x 10¹⁴ electrons

2. F2 = 2.4 N

8. number of excess electrons = 3.8 x 10¹⁴ electrons, 3.4 x 10^-16 kg increase (since you know the increase of electrons, and the mass of one electron is 9.11 x 10^-31 kg)

11. the three net forces on each F1 = 1.2 x 10²N to the left, F2 = 5.3 x 10²N to the right, F3 = 3.9 x 10² N.

20. If one charge is Q1 , the other charge will be Q2 = Q – Q1 . For the force to be repulsive, the two charges must have the same sign. Because the total charge is positive, each charge will be positive. We account for this by considering the force to be positive:

F = kQ1Q2/r2 = kQ1(Q – Q1)/r2;

You know F = 12 N and r = 1.06 meters. This gives you a quadratic equation that can be solved numerically (on the calculator!) for Q1 = 50.0 x 10^-6 C, Q2 = 30 x 10^-6 C